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For what values of m does the function y = $Ae^{mt}$ satisfy the following equation? $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0$

I tried taking the first and second derivative of the function, but I got stuck there. $\frac{dy}{dx} = Ate^{mt}$

$\frac{d^2y}{dx^2} = At^2e^{mt}$

$-6y = -6Ae^{mt}$

Adam Staples
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Amandha
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2 Answers2

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Let's find $y'$ and $y''$.

$y=Ae^{mx}$ is given.

$\frac{dy}{dx}=A\cdot me^{mx}$.

$\frac{d^2y}{dx^2}=A\cdot m\cdot me^{mx}=Am^2e^{mx}$.

Now we plug these into the differential equation above.

$Am^2e^{mx}+Ame^{mx}-6Ae^{mx}=0\\ \implies Ae^{mx}(m^2+m-6)=0\\ \implies Ae^{mx}(m+3)(m-2)=0$

So what values of $m$ would make the last equation equal to $0$?

Well, $Ae^{mx}$ can't be zero (unless $A=0$), which just leaves:

$m+3=0\implies m=-3$ and $m-2=0\implies m=2$.

Hence the values of $m$ which would satisfy the differential equation are $m=2$ or $m=-3$.

Sujaan Kunalan
  • 10,944
  • It's wrong, if he indeed meant $y=Ae^{mt}$. However, if he meant $y=Ae^{mx}$, and the $t$ was a typo, this is right. –  Apr 05 '14 at 16:42
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If you did indeed mean $y=Ae^{mx}$, then we can substitute $$\dfrac{dy}{dx}=Ame^{mx}\\ \implies \dfrac{d^2y}{dx^2}=Am^2e^{mx}$$ Substituting gives $$Am^2e^{mx}+Ame^{mx}-6Ae^{mx}=0\\ \implies\text{ For nonzero $A$, }m^2+m-6=0\\ \implies m=\dfrac{-1\pm\sqrt{25}}{2}$$